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1 year ago

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You place a 500 g block of an unknown substance in an insulated container filled 2 kg of water. The block has an initial tempera

ture of 50 degrees C. The water is initially at 20 degrees C. If the equilibrium temperature of the block and water is 25 degrees C, what is the specific heat of the block? The specific heat of water is 4186 J. kg K. 3349 J, kg C 1189 J, kg C 5545 J, kg C 750 J/kg C 2080 J/kg C

1 answer:

Nina [5.8K]1 year ago

4 0

Answer:

3349J/kgC

Explanation:

Questions like these are properly handled having this fact in mind;

Heat loss = Heat gained

Quantity of heat = mcΔ∅

m = mass of subatance

c = specific heat capacity

Δ∅ = change in temperature

m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)

m₁ = mass of block = 500g = 0.5kg

c₁ = specific heat capacity of unknown substance

∅₂ = block initial temperature = 50oC

∅₁ = equilibrium temperature of block and water after mix= 25oC

m₂= mass of water = 2kg

c₂ = specific heat capacity of water = 4186J/kg C

∅₃ = intial temperature of water = 20oC

0.5c₁(50-25) = 2 x 4186(25-20)

And we can find c₁ which is the unknown specific heat capacity

c₁ = $\frac{2*4186*5}{0.5*25}$ = 3348.8J/kg C≅ 3349J/kg C

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In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce

noname [10]

Answer:

14.7 m/s

Explanation:

a = acceleration experienced by driver's head = 50 g = 50 x 9.8 m/s² = 490 m/s²

v₀ = initial speed of the driver = 0 m/s

v = final speed of the driver after 30 ms

t = time interval for which the acceleration is experienced = 30 ms = 0.030 s

Using the equation

v = v₀ + a t

Inserting the values

v = 0 + (490) (0.030)

v = 14.7 m/s

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2 years ago

If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he

shutvik [7]

Answer: 9938.8 km

Explanation:

1 pound-force = 4.48 N

30.0 pounds-force = 134.4 N

The force of gravitation between Earth and object on the surface of is given by:

$F = \frac{GMm}{R^2} = mg$

Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.

At height, h above the surface of the Earth, the weight of the object:

$(mg)'= \frac{GMm}{(R+h)^2}$

we need to find "h"

taking the ratio of two:

$\frac{mg}{(mg)'}=\frac{(R+h)^2}{R^2}\\ \Rightarrow \frac{90kg \times 9.8 m/s^2}{134.4 N}=\frac{(R+h)^2}{R^2}\\ \Rightarrow 6.56 R^2= (R+h)^2 \Rightarrow h= (2.56-1)R\\ \Rightarrow h = 1.56 R = 1.56 \times 6371 km = 9938. 8 km$

Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.

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2 years ago

A 500 kg motorcycle accelerates at a rate of 2 m/s .how much force was applied to the motorcycle?

Aleksandr [31]

Answer:

by using formula F=ma which is m stand for mass a stand for acceleration. so 500kg × 2 ms^-2

8 0

1 year ago

Read 2 more answers

he drawing shows two perpendicular, long, straight wires, both of which lie in the plane of the paper. The current in each of th

AleksandrR [38]

Answer:

The magnitudes of the net magnetic fields at points A and B is 2.66 x $10^{-6}$ T

Explanation:

Given information :

The current of each wires, I = 4.7 A

dH = 0.19 m

dV = 0.41 m

The magnetic of straight-current wire :

B= μ $_{0}$ I/2πr

where

B = magnetic field (T)

μ $_{0}$ = 1.26 x $10^{-6}$ (N/ $A^{2}$ )

I = Current (A)

r = radius (m)

the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,

BH = μ $_{0}$ I/2πr

= (1.26 x $10^{-6}$ )(4.7)/(2π)(0.19)

= 4.96 x $10^{-6}$ T

BV = μ $_{0}$ I/2πr

= (1.26 x $10^{-6}$ )(4.7)/(2π)(0.41)

= 2.3 x $10^{-6}$ T

hence,

the net magnetic field = BH - BV

= 4.96 x $10^{-6}$ - 2.3 x $10^{-6}$

= 2.66 x $10^{-6}$ T

4 0

1 year ago

A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnet

Mumz [18]

<h3>Question:</h3>

A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.

<h3>Answer:</h3>

1.6nT [in the negative z direction]

<h2>Explanation:</h2>

The magnetic field, B, due to a distance of finite value b, is given by;

B = (μ₀IL) / (4πb $\sqrt{b^2 + L^2}$ ) -----------(i)

Where;

I = current on the wire

L = length of the wire

μ₀ = magnetic constant = 4π × 10⁻⁷ H/m

From the question,

I = 20A

L = 2.0cm = 0.02m

b = 5.0m

Substitute the necessary values into equation (i)

B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 $\sqrt{5.0^2 + 0.02^2}$ )

B = (10⁻⁷ x 20 x 0.02) / (5.0 $\sqrt{5.0^2 + 0.02^2}$ )

B = (10⁻⁷ x 20 x 0.02) / (5.0 $\sqrt{25.0004}$ )

B = (10⁻⁷ x 20 x 0.02) / (25.0)

B = 1.6 x 10⁻⁹T

B = 1.6nT

Therefore, the magnetic field at the point x = 5.0m on the x-axis is 1.6nT.

PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.

5 0

2 years ago